Question: Simplify and expand the following expression: $ \dfrac{1}{5r - 15}+ \dfrac{4}{2r - 14}+ \dfrac{4}{r^2 - 10r + 21} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{1}{5r - 15} = \dfrac{1}{5(r - 3)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{4}{2r - 14} = \dfrac{4}{2(r - 7)}$ We can factor the quadratic in the third term: $ \dfrac{4}{r^2 - 10r + 21} = \dfrac{4}{(r - 3)(r - 7)}$ Now we have: $ \dfrac{1}{5(r - 3)}+ \dfrac{4}{2(r - 7)}+ \dfrac{4}{(r - 3)(r - 7)} $ The least common multiple of the denominators is: $ 10(r - 3)(r - 7)$ In order to get the first term over $10(r - 3)(r - 7)$ , multiply by $\dfrac{2(r - 7)}{2(r - 7)}$ $ \dfrac{1}{5(r - 3)} \times \dfrac{2(r - 7)}{2(r - 7)} = \dfrac{2(r - 7)}{10(r - 3)(r - 7)} $ In order to get the second term over $10(r - 3)(r - 7)$ , multiply by $\dfrac{5(r - 3)}{5(r - 3)}$ $ \dfrac{4}{2(r - 7)} \times \dfrac{5(r - 3)}{5(r - 3)} = \dfrac{20(r - 3)}{10(r - 3)(r - 7)} $ In order to get the third term over $10(r - 3)(r - 7)$ , multiply by $\dfrac{10}{10}$ $ \dfrac{4}{(r - 3)(r - 7)} \times \dfrac{10}{10} = \dfrac{40}{10(r - 3)(r - 7)} $ Now we have: $ \dfrac{2(r - 7)}{10(r - 3)(r - 7)} + \dfrac{20(r - 3)}{10(r - 3)(r - 7)} + \dfrac{40}{10(r - 3)(r - 7)} $ $ = \dfrac{ 2(r - 7) + 20(r - 3) + 40} {10(r - 3)(r - 7)} $ Expand: $ = \dfrac{2r - 14 + 20r - 60 + 40}{10r^2 - 100r + 210} $ $ = \dfrac{22r - 34}{10r^2 - 100r + 210}$ Simplify: $ = \dfrac{11r - 17}{5r^2 - 50r + 105}$